Problem: In rectangle $ABCD$, $AB = 6$ cm, $BC = 8$ cm, and $DE = DF$. The area of triangle $DEF$ is one-fourth the area of rectangle $ABCD$. What is the length in centimeters of segment $EF$? Express your answer in simplest radical form.

[asy]
draw((0,0)--(0,24)--(32,24)--(32,0)--cycle);
draw((13,24)--(32,5));
label("$A$",(0,24),W);
label("$B$",(0,0),W);
label("$C$",(32,0),E);
label("$D$",(32,24),E);
label("$E$",(13,24),N);
label("$F$",(32,5),E);
[/asy]
Solution: The area of the rectangle is $(6)(8)=48$, so the area of triangle $DEF$ is $48/4 =12$.  Since $DE=DF$, the area of $DEF$ is $(DE)(DF)/2 = DE^2/2$, so $DE^2/2 = 12$.  Therefore, $DE^2 = 24$.  From the Pythagorean Theorem, we have  \[EF^2 = DE^2 +DF^2 = 24+24=48,\] so $EF =\sqrt{48} = \boxed{4\sqrt{3}}$.